The Gaseous State

The Gaseous StateChapter 511AgendaPreliminariesProperties of GasesGas LawsIdeal Gas LawDaltons LawKinetic Molecular TheoryReal Gases

2Coverage in Tro et al.5.2 Pressure: The Result of Molecular Collisions (exclude manometers)5.3 The Simple Gas Laws5.4 The Ideal Gas Law5.5 Applications of the Ideal Gas Law5.6 Mixtures of Gases and Partial Pressures5.7 Gases in Chemical Reactions: Stoichiometry Revisited5.8 Kinetic Molecular Theory: A Model for Gases5.9 Mean Free Path, Diffusion, and Effusion of Gases5.10 Real Gases3Preliminaries4Pressure concept at molec levelMomentum transfer:Particle p = -px -px = -2px wall p = +2pxForce:over a time interval, t, momentum, , is transferred to the wall from large number of collisionsF = /tPressure:P = F/A5pxpyp-pxpyp5Pressure unitsPressure = force per unit areaSI Unit1 Pa = 1 Nm-2 = 1 kgm-1.s-2Related units1 kPa = 1000 Pa1 bar = 105 Pa (exactly)1 atm = 101,325 Pa6

6Other pressure unitstorr (name) or Torr (unit)Pressure measured with barometerAtmospheric P pushes Hg up tubeHg level rises to point where P due to Hg weight balances atmos. PMeasure height (h) of Hg columnP = gh1 Torr = P required to raise Hg column by 1 mm1 bar = 750.0 Torr7

7LC: Pressure UnitsThe normal atmospheric pressure at the top of Mt Everest (8,848 m) is about 0.308 atm. What is that in bar if 1 atm = 101,325 Pa?0.3020.3120.3220.33288LC: Pressure Units99Some Terms To MemorizeStandard temperature and pressure (STP):0C (273.15 K) and 1 bar

Room temperature and pressure (RTP):25C (298.15 K) and 1 bar

Know these terms (there are only two)1010Properties of Gases11Properties of GasesExpand to fill the volume of the containerHighly compressibleNo phase separation in mixturesMolecules only weakly interactGaseous substances at RTP:Some molecular elements (e.g. Ar, Cl2, F2, N2, etc.)Many small molecules (e.g. HCl, NH3)1212

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1414The Gas Laws15Boyles LawThe volume of a gas maintained at constant temperature is inversely proportional to the pressurei.e., at constant moles and T,V = C/P where C is a constantOR:PfVf = PiVi1616Boyles Experiment:17

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1818Charless LawThe volume of a gas maintained at constant pressure is directly proportional to the absolute temperature (in kelvins)

V = CT (constant moles and P)or

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2020

2121Blue balloon dipped in liquid nitrogen22

22LC: Charless LawA helium balloon with a volume of 2.00 L at 22.0C is dropped into a large container of liquid nitrogen (196.0C). The final volume of the gas in the balloon is:

0.224 L0.522 L7.65 L17.8 L2323Solution: Charless LawA helium balloon with a volume of 2.00 L at 22.0C is dropped into a large container of liquid nitrogen (196.0C). The final volume of the gas in the balloon is:Vi = 2.00 LTi = (22.0 + 273.15) K = 295.2 KTf = (-196.0 + 273.15) K = 77.2 K24

24Avogadros LawEqual volumes of gases at the same temperature and pressure contain equal numbers of moleculesV = const. n

At 25C and 1 bar pressure, V/n 25 L/mol2525

2626Ideal Gas LawPutting it all together27Ideal Gas LawCombines all the gas laws:PV = nRTR is the gas constant:R = 8.314 J K-1mol-1 = 8.3145 Pa m3 mol-1 K-1= 0.08314 L bar mol-1 K-1(= 0.08206 L atm mol-1 K-1)So, PV = const. at constant T, n (Boyles Law)V T at constant P, n (Charless Law)V n at constant T, P (Avogadros Law)

2828Pictures and realityO2 collision diameter (c.d.) = 3.467 Window volume: 6 c.d. wide 1 c.d. deep r2d = 1178 3 V = 1.17810-27 m3 Density: 8.5/1178 3 or 7.21027 m-3 n/V = 1.2104 molm-3Pressure at 298K: ~300 bar!29

Example3030Example3131LC: Ideal Gas LawAn empty aerosol can has a volume of 0.406 L and contains 0.025 mol of a propellant gas. What is the pressure of the gas at 25.0C? (R=0.083145 Lbarmol1K1)0.13 bar1.3 bar1.5 bar1.5102 bar3232Solution: Ideal Gas Law3333Ideal Gas Law (IGL) ProblemsUsually at least one variable fixedFixed variables are your handleStart by figuring out what is fixed and what is changing3434Example3.29 mol of ideal gas are contained inside a cylinder-and-piston system under an external pressure 1.01 bar, which is balanced by the pressure of the ideal gas. Calculate the volume change as the system is heated from 300 K to 500 K. (R=0.083145 L bar/mol/K)What is changing?Volume (to find) and temperature (300-500 K)What is fixed?Moles (sealed by piston) & pressure (1.01 bar)Suggests Charless Law3535Apparatus361.01 bar1.01 bar300 K500 K36AnswerV = Vf - ViVi = nRTi/P (one unknown) = (3.29 mol)(0.083145 L bar/mol/K)(300 K) (1.01 bar) = 81.2 Lby Charless Law, Vf = Vi(Tf/Ti)Vf = (81.2 L)(500 K/300 K) = 135.4 LV = Vf - Vi = 135.4 L - 81.2 L = 54 L3737Using the handleA weather balloon ascends to an altitude at which T=226 K and P=0.26 bar. If the balloon had an initial volume of 7.2 L at sea level (T=288 K; P=1.00 bar), what is the volume at the high altitude? (R=0.083145 L bar/mol/K)9.6 L14 L22 L30L

3838Example 2What is changing?T (293 to 250K), V (7.2 L to ?) and P (1.00 to 0.45 bar)What is constant?Moles (sealed in a baloon)ni = nfPiVi/(RTi) = PfVf/(RTf) Vf = ViPiTf/(PfTi)Vf = (7.2 L)(1.00 bar)(226 K) (0.26 bar)(288 K)Vf = 22 L (Answer C)3939Application: Finding Molar MassA 536-mL glass bulb is filled with a pure gas at 294 K and 1.01 bar pressure. The mass of the gas is 1.59 g. What is the molar mass? (R=0.083145 L bar/mol/K)

How can we relate molar mass (M) to PV=nRT?Through the moles (n)M = m/n (molar mass is mass of one mole of substance)We can substitute n = m/M

4040Application: Finding Molar MassPV = nRT and n=m/M PV = mRT/MRearranging for M:M = mRT/(PV)M = (1.59 g)(0.083145 Lbar/mol/K)(294 K)(1.01 atm)(0.536 L)= 71.8 g mol-1

4141LC: Finding Molar MassA 25.0-mL glass bulb is filled with a pure gas at 25C and 380 Torr pressure. The mass of the gas is 42.8 mg. What is the molar mass?(R = 0.083145 L bar/mol/K; 1 bar = 750.0 Torr)

41.9 g/mol 83.7 g/mol167 g/mol7.02103 g/mol

4242Solution: Finding Molar MassA 25.0-mL glass bulb is filled with a pure gas at 25C and 380 Torr pressure. The mass of the gas is 42.8 mg. What is the molar mass?P = 0.507 bar; T = 298 K; V=0.025 L; m =0.0428 g

M = (0.0428 g)(0.083145 L bar/mol/K)(298 K)(0.507 bar)(0.0250 L)= 83.7 g mol-14343Finding M from Gas DensityIn the previous example we substituted the mass and molar mass for n to getM = mRT/(PV)

Note that m/V = (rho = the density of the gas) M = RT/P

Molar mass data can be converted to density data and vice versa4444ExampleAir is a mixture of N2, O2, Ar, and other gases; therefore, it does not have a well-defined molar mass. However, the average molar mass of air is approximately 29.0 g/mol. What is the density of air at room temperature and pressure? (R=0.083145 L bar/mol/K) = MP/(RT)= (29.0 g/mol)(1.00 bar) .(0.083145 L bar mol-1 K-1)(298 K)= 1.19 g/L4545Daltons LawMixing it up46Daltons Law of Partial PressuresP of a gas mixture = sum of the pressures that the components would exert individually if separated.For a mixture of N components, P = P1 + P2 + P3 + + PNFor ideal gases, P1 = n1RT/V, etc. P = (n1+n2+n3++nN)RT/V47

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4949ExampleWhat is the P of 10.0 g of O2 and 6.00 g of C2H4 in a 12.0 L bulb at 298 K? (MO2 = 32.0 g/mol; MC2H4 = 28.1 g/mol)ANS: nO2 = (10.0 g)/(32.0 g/mol) = 0.312 molnC2H4 = (6.00 g)/(28.1 g/mol) = 0.213 mol

P = (nO2 + nC2H4)RT/VP = (0.312 + 0.213)(0.083145)(298)/(12.0)= 1.08 bar5050LC: Partial PressuresWhat is the total pressure of a mixture of 1.570 mol of CO and 0.870 mol NO2 in a 25.0 L container at 25C?(R=0.083145 L bar/mol/K; 0C = 273.15 K)

0.203 bar1.54 bar2.42 bar242 bar5151Solution: Partial Pressures5252Mole fractionPartial pressure of gas A can be expressed in terms of the mole fraction of A, A:PA = APtotalFrom Daltons law:Ptotal = (PA + PB ++ PN) = ntotalRT/Vand PA = nART/VsoPA/Ptotal = nA/ntotal = A

53Collecting Gases Over Water54

AND water vapourca. 20 mbardepends on temp.54LC: Gases Over Water A 41.4 mL sample of a unknown pure gas is collected over water at 25.0C and 769 Torr atmospheric pressure. What is the partial pressure of the unknown gas? (PH2O = 23.77 Torr; 1 bar = 750.0 Torr)

0.032 bar0.994 bar1.03 bar1.14 bar5555Solution: Gases Over Water A 41.4 mL sample of a unknown pure gas is collected over water at 25.0C and 769 Torr atmospheric pressure. What is the partial pressure of the unknown gas? (PH2O = 23.77 Torr; 1 bar = 750.0 Torr)

ANSWER: B

Pgas = (769 Torr - 23.77 Torr)/750 Torrbar-1= 0.994 bar5656A better exampleA sample containing CaCO3 is decomposed under high temperature, producing CO2 which is collected over water. The volume of gas collected is 34.1 mL at 22C and 756 Torr atmospheric pressure. How many moles of CO2 were produced? What mass of CaCO3 (100.1 g/mol) was present in the original sample? PH2O = 19.8 Torr1 bar = 750.0 TorrR = 0.083145 Lbarmol-1K-15757Answer (part 1)By Daltons Law: P = PH2O + PCO2 = 756 TorrPH2O = 19.8 Torr PCO2 = 736 Torr750 torr = 1 bar PCO2 = 0.981 bar

nCO2 = PCO2V/(RT)= (0.981 bar)(34.110-3 L) . (0.083145 Lbarmol-1K-1)(295 K)= 1.3510-3 mol5858Answer (part 2)Need a balanced chemical equation:CaCO3 (s) CaO (s) + CO2 (g)

So, 1 mol CaCO3 produces 1 mol CO2nCaCO3 = nCO2 (1 mol CaCO3/1 mol CO2)= 1.36 10-3 mol

mCaCO3 = (1.36 10-3 mol)(100.1 g/mol) = 0.136 g5959Kinetic Molecular TheoryHow a molecular model explains gas behaviour60Postulates of the theoryGases modelled as a collection of moleculesGas molecules are in continuous, random motionVolume of gas >> volume of molecules Average kinetic energy is proportional to temperatureDoes not depend on chemical identityCollisions are elasticIntermolecular forces are negligible6161Molecular Picture of Gas PressureMomentum transfer:Particle p = -px -px = -2px wall p = +2pxForce:over a time interval, t, momentum, , is transferred to the wall from large number of collisionsF = /tPressure:P = F/A62pxpyp-pxpyp62Gas Laws ExplainedBoyles LawIf V (at const. T), collision frequency , so P Charless LawIf T , collision frequency , so P but P is constant so V insteadAvogadros LawIf n (at const. P & T), collision frequency , so P Daltons LawMolecules do not interactAverage energy of molecule depends solely on T

63Ideal gas lawCan be derived (roughly) using postulatesYou can look it up in the book if you are interested

64Molecular SpeedsWe will cover molecular speeds in the following slidesWill not cover equation 5.26Will not cover section 5.9 MFP, Diffusion, EffusionYou will be expected to interpret the graphs that follow65Molecular Speeds and Temp66

66Molecular Speeds and Mass67

67LC: Kinetic Molecular TheoryWhich of the following second period gases would have molecules traveling at the slowest average speed at 25C?

N2O2F2All these gases will have the same average speed6868LC: Kinetic Molecular TheoryLets imagine that you have a magic radar gun that can clock a single molecule in a gas mixture at 25C. You pick one N2 molecule, one O2 molecule and one F2 molecule from the mixture and measure their speeds. Which has the highest speed?N2O2F2Theyre all the sameNot enough information6969Real GasesAdding molecular volumes and attractions to the theory70Real GasesIdeal gases are theoretical:Molecules occupy no spaceMolecules do not attract one anotherFor ideal gases,

Real gases approximate ideality at high T & low P71

71Real Gas BehaviourNegative deviations: attractive forcesPositive deviations: molecular volumes72

72Effects of Temperature on N2 Gas73

73Van der Waals Gas Equation74Van der Waals Example75Van der Waals Example76SummaryKnow your gas laws (esp. IGL)Applications & extensions of IGLCollecting gases over waterFinding the molar massKinetic Molecular Theory Know and understand itBe able to interpret graphs of speed distributionsReal gases deviate from ideal behaviour because ofmolecular interactionsmolecular volumesKnow vdW equation & how to interpret non-ideal graphs7777PracticeReview questions 1, 4, 6, 8-14, 17-24. Problems by topic 25, 31, 33, 35, 37, 45, 49, 51, 53, 55, 57, 61, 65, 73, 77, 79, 83, 85, 87, 91, 93, 99, 119, 129, 135.

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